You have found the following ages (in years) of 6 zebras. Those zebras were randomly selected from the 30 zebras at your local zoo: $ 22,\enspace 21,\enspace 13,\enspace 19,\enspace 16,\enspace 18$ Based on your sample, what is the average age of the zebras? What is the standard deviation? You may round your answers to the nearest tenth.
Because we only have data for a small sample of the 30 zebras, we are only able to estimate the population mean and standard deviation by finding the sample mean $({\overline{x}})$ and sample standard deviation $({s})$ To find the sample mean , add up the values of all $6$ samples and divide by $6$ $ {\overline{x}} = \dfrac{\sum\limits_{i=1}^{{n}} x_i}{{n}} = \dfrac{\sum\limits_{i=1}^{{6}} x_i}{{6}} $ $ {\overline{x}} = \dfrac{22 + 21 + 13 + 19 + 16 + 18}{{6}} = {18.2\text{ years old}} $ Find the squared deviations from the mean for each sample. Since we don't know the population mean, estimate the mean by using the sample mean we just calculated {14.44} + {7.84} + {27.04} + {0.64} + {4.84} + {0.04}} {{6 - 1}} $ {s^2} = \dfrac{{54.84}}{{5}} = {10.97\text{ years}^2} $ As you might guess from the notation, the sample standard deviation $({s})$ is found by taking the square root of the sample variance $({s^2})$ ${s} = \sqrt{{s^2}}$ $ {s} = \sqrt{{10.97\text{ years}^2}} = {3.3\text{ years}} $ We can estimate that the average zebra at the zoo is 18.2 years old. There is also a standard deviation of 3.3 years.